Ideal correspondence

I'm confusing the ideal correspondence theorem. Is the following right?

Ideal correspondence: Let $f:A \to B$ be a ring homomorphism. Then there is a one-to-one order-preserving correspondence between ideals of $f(A)$ and ideals of $A$ which contain $\ker(f)$ by $\alpha \mapsto f(\alpha)$ (conversely $\beta \mapsto f^{-1}(\beta)$), and prime ideals correspond to prime ideals, and maximal ideals correspond to maximal ideals.

Solution 1:

Yes. For me, everything is much less confusing if you think of ideals $I$ in terms of the corresponding quotient maps $A \to A/I$. Then the ideals of $f(A)$ are precisely the quotient maps $f(A) \to f(A)/I$, which composed with the map $A \to f(A)$ are precisely those maps $A \to f(A) \to f(A)/I$ which factor through $f(A)$ - in other words, which have kernel containing $\ker(f)$.

Moreover, an ideal $I$ is prime if and only if $A/I$ is an integral domain and maximal if and only if $A/I$ is a field. Hence if $f(A) \to f(A)/I$ has image an integral domain then the composition $A \to f(A) \to f(A)/I$ also has image an integral domain, and the same is true if you replace "integral domain" by "field." Other properties of ideals can also be stated in this way, hence also pull back: for example, an ideal $I$ is radical if and only if $A/I$ has no nilpotents.

Solution 2:

Yes. You might make the direction of correspondence consistent by saying that ideals of $A$ which contain ker($f$) go to ideals of $f(A)$ under $\alpha$ going to $f(\alpha)$, but you clearly have the right idea. In this mapping prime ideals that contain ker($f$) go to prime ideals of $f(A)$, and so also with maximal ideals containing ker($f$) going to maximal ideals of $f(A)$. The statement could be a bit more concise if we require $f$ onto, so that $f(A)$ is replaced in the the statement by $B$.